Application of Derivatives 2 Question 27
####27. Using the relation $2(1-\cos x)<x^{2}, x \neq 0$ or prove that $\sin (\tan x) \geq x, \forall x \in[0, \pi / 4]$.
(2003, 4M)
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Answer:
Correct Answer: 27. (c)
Solution:
- Let
$ \begin{aligned} f(x) & =\sin (\tan x)-x \\ f^{\prime}(x) & =\cos (\tan x) \cdot \sec ^{2} x-1 \\ & =\cos (\tan x)\left(1+\tan ^{2} x\right)-1 \end{aligned} $
$=\tan ^{2} x{\cos (\tan x)}+\cos (\tan x)-1 $
$>\tan ^{2} x \cos (\tan x)-\frac{\tan ^{2} x}{2} $
$\because 2(1-\cos x)<x^{2}, x \neq 0 \Rightarrow \cos x>1-\frac{x^{2}}{2} $
$\Rightarrow \quad \cos (\tan x)>1-\frac{\tan ^{2} x}{2} $
$f^{\prime}(x)>\tan ^{2} x \cos (\tan x)-\frac{1}{2}$
$>\tan ^{2} x[\cos (\tan x)-\cos (\pi / 3)]>0$
$\Rightarrow f(x)$ is increasing function, for all $x \in[0, \pi / 4]$
As
$ \begin{aligned} f(0)=0 \Rightarrow & f(x) \geq 0, \text { for all } x \in[0, \pi / 4] \\ & \sin (\tan x) \geq x \end{aligned} $
$ \Rightarrow $
